If you have a non-scientific calculator you can use the Newton-Raphson method.
Suppose you wish to find the square root of 7.
Let f(x) = x2 - 7 so that f(x) = 0 when x is the square roo. That is, you want to find x such that f(x) = 0.
Let f'(x) = 2x
[f'(x) is the derivative of f(x) but you do not need to know that to use the N-R method.]
Make a guess at the square root of 7, and call is x0.
Then calculate
xn+1 = xn - f(xn)/f'(xn) for n = 1, 2, 3, ...
Provided you made a reasonable choice for the starting point, the iteration will very quickly converge to the true answer. Even if your first guess is not so good:
Suppose you start with x0 = 5 (a pretty poor choice since 52 is 25, which is nowhere near 7).
Even so, x3 = 2.2362512515, which is less than 0.01% from the true value. Finally, remember that the negative value is also a square root.
1.7720045 is the approximate square root of 3.14.
7
approx ±49*i where i is the imaginary square root of -1.
Because if you approximate pi, you get pi = 3.1415926535897932384626433832795...Square root of 3.1415926535897932384626433832795... is 1.7724538509055160272981674833411...... which is the square root of pi.Hope that helps.
The approximate square root of 150 is between 12 and 13. It is 12.2!
1.7720045 is the approximate square root of 3.14.
The square root of 79 is approximately 8.89. Rounding to the nearest whole number, the approximate square root of 79 is 9.
7
10.954i
approx ±49*i where i is the imaginary square root of -1.
Because if you approximate pi, you get pi = 3.1415926535897932384626433832795...Square root of 3.1415926535897932384626433832795... is 1.7724538509055160272981674833411...... which is the square root of pi.Hope that helps.
The approximate square root of 150 is between 12 and 13. It is 12.2!
Approx 9.592
It is approximately 5.1769
5.2915 is. (rounded)
1,290 is not a perfect square. The approximate square root is ± 35.916569992136
√67 = approximately 8.2